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Song Classification for Dancing

Explore a new notion of pattern matching tailored for musical sequences to analyze rhythms with varying event durations. Discover a novel approach to identifying extended musical patterns in sequences.

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Song Classification for Dancing

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  1. Song Classification for Dancing Manolis Cristodoukalis, Costas Iliopoulos, M. Sohel Rahman, W.F. Smyth PSC 2006

  2. Classical Pattern Matching • Input: A text T = T[1..n], A Pattern P = P[1..m], both over the alphabet . • Output: • Whether P occurs in T • If yes, then the set occ(P) = {i | T[i..i+m-1] = P} PSC 2006

  3. Example – Pattern matching Pattern P = AAGCTA Text T = CAAGCTAAGCTAC A A G C T A Pattern A A G C T A 1 2 3 4 5 6 7 8 9 10 11 12 13 Text C A A G C T A A G C T A C Occ(P) = {2, 7} PSC 2006

  4. Our Case • Input: • Text A musical Sequence • Pattern Rhythm • Output: • Slightly different from Classical PM • A ‘extended/modified’ notion of PM PSC 2006

  5. Musical Sequence • A string t = t[1]t[2]…t[n] • t[i]  N+ • Example: • [0,50,100,200,250,300,350,400,500,550] • Significance: Sequence of events! PSC 2006

  6. Musical Sequence • A string t = t[1]t[2]…t[n] • t[i]  N+ • Example: • [0,50,100,200,250,300,350,400,500,550] • Some (musical) event occurs at 0 ms PSC 2006

  7. Musical Sequence • A string t = t[1]t[2]…t[n] • t[i]  N+ • Example: • [0,50,100,200,250,300,350,400,500,550] • Some (musical) event occurs at 0 ms • Some (musical) event occurs at 50 ms PSC 2006

  8. Musical Sequence • A string t = t[1]t[2]…t[n] • t[i]  N+ • Example: • [0,50,100,200,250,300,350,400,500,550] • Some (musical) event occurs at 0 ms • Some (musical) event occurs at 50 ms • Some (musical) event occurs at 100 ms • And so on… PSC 2006

  9. Musical Sequence • [0,50,100,200,250,300,350,400,500,550] • Alternative Representation (We use) • [50,50,100,50,50,50,50,100,50] • Significance: Sequence of “duration” of theevents! PSC 2006

  10. Musical Sequence First Representation 0 50 100 200 250 300 350 450 500 550 50 Duration of First Event = 50 – 0 = 50 PSC 2006

  11. Musical Sequence First Representation 0 50 100 200 250 300 350 450 500 550 50 50 Duration of Second Event = 100 – 50 = 50 PSC 2006

  12. Musical Sequence First Representation 0 50 100 200 250 300 350 450 500 550 50 50 100 Duration of Second Event = 200 – 100 = 100 PSC 2006

  13. Musical Sequence First Representation 0 50 100 200 250 300 350 400 500 550 50 50 100 50 50 50 50 100 50 Alternative Representation PSC 2006

  14. Rhythm • A string r = r[1]r[2]…r[m] • r[j]  {Q, S} • Q means a quick(er) event • S means a slow(er) event • S is double the length (duration) of Q • Exact length of Q or S is not a priori known PSC 2006

  15. New Notion of Match • Let Q q  N+ • Then S  2  q • Q matches t[i..i’] iff • q = t[i] + t[i+1] + … + t[i’] • 1  i  i’  n • If i = i’ then the match is SOLID PSC 2006

  16. New Notion of Match Let q = 150 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50 Q 50+100 = 150 = q PSC 2006

  17. New Notion of Match Let q = 150 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50 Q Q 100 + 50 = 150 = q PSC 2006

  18. New Notion of Match Let q = 150 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50 Q Q Q Q Q Q PSC 2006

  19. New Notion of Match Let q = 100 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50 Q Q Q Q Q Q Solid Match PSC 2006

  20. New Notion of Match • S  2  q • S matches t[i..i’] iff either of following is true: • i = i’ and t[i] = 2q (SOLID) • i  i’ and there exists i  i1  i’ such that • q = t[i] + t[i+1] + … + t[i1] = t[i1+1] + t[i+1] + … + t[i’] • 1  i  i’  n • So, S is either solid or a tile of 2 consecutive Q’s PSC 2006

  21. New Notion of Match Let q = 100, then S  2q = 200 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50 S t[1] + t[2] = 100 = t[3] PSC 2006

  22. New Notion of Match Let q = 100, then S  2q = 200 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50 S S t[4] + t[5] = 100 = t[6] + t[7] PSC 2006

  23. New Notion of Match Let q = 100, then S  2q = 200 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50 S S S S PSC 2006

  24. New Notion of Match Let q = 100, then S  2q = 200 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50 S S S S S Although, t[2] + t[3] + t[4] = 200 = 2q PSC 2006

  25. New Notion of Match Let q = 100, then S  2q = 200 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50 S S S S S S PSC 2006

  26. New Notion of Match Let r = QSS and Let q = 50 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50 Q S S r r matches t[2..5] PSC 2006

  27. New Notion of Match Let r = QSS and Let q = 50 1 2 3 4 5 6 7 8 9 Q S 50 50 100 50 50 50 50 100 50 S Q S S r r r matches t[5..8] PSC 2006

  28. New Notion of Match Let r = QSS and Let q = 50 1 2 3 4 5 6 7 8 9 Q S 50 50 100 50 50 50 50 100 50 S Q S S r r r covers t[2..8] PSC 2006

  29. Our Problem • Input: • A musical Sequence t • A rhythm r • Output: • The longest substring t[i..i’] that is covered by r PSC 2006

  30. Restriction • For each match of r, at least one S must be solid Let r = QSS and Let q = 50 1 2 3 4 5 6 7 8 9 r 50 50 100 50 50 50 50 50 100 Q S S SOLID S PSC 2006

  31. Restriction • For each match of r, at least one S must be solid Let r = QSS and Let q = 50 1 2 3 4 5 6 7 8 9 r 50 50 100 50 50 50 50 50 100 Q S S r At least one S must be Solid PSC 2006

  32. Restriction • For each match of r, at least one S must be solid Let r = QSS and Let q = 50 1 2 3 4 5 6 7 8 9 r 50 50 100 50 50 50 50 50 100 Q S SOLID S S At least one S must be Solid r PSC 2006

  33. Motivation • Our aim is to classify music according to dancing rhythm • Music seq can be considered as a series of events corresponding to music signals • Drum beats • Guiter picks • Horn hits PSC 2006

  34. Motivation • The intervals between these events characterize how the song is danced • Basically two dancing rhythms: Quick & Slow • Example: • cha-cha  SSQQSSSQQS • foxtrot  SSQQSSQQ • jive  SSQQSQQS PSC 2006

  35. Motivation • So solution to our problem can classify songs according to dancing rhythms! PSC 2006

  36. Algorithm • Stage 1: Find all occurrences of S for a chosen value    such that /2  . • Stage 2: Transform areas around each S into sequences of Q. • Stage 3: Find the matches of r and consequently the cover. PSC 2006

  37. Algorithm: Stage 1 • We construct two arrays: • first[1..||]: first[] = i iff the first occurrence of symbol  appears at position i. • next[1..n]: next[i] = j iff the next occurrence of symbol at t[i] appears at t[j]. PSC 2006

  38. Algorithm: Stage 1 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50  = {50, 100}, assume indexed alphabet PSC 2006

  39. Algorithm: Stage 1 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50  = {50, 100}, assume indexed alphabet First occurrence of 50 first[1] = 1 PSC 2006

  40. Algorithm: Stage 1 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50  = {50, 100}, assume indexed alphabet first[1] = 1 first[2] = 3 First occurrence of 100 PSC 2006

  41. Algorithm: Stage 1 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50 next[1] = 2 Next occurrence of t[1] = 50 PSC 2006

  42. Algorithm: Stage 1 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50 next[1] = 2 next[2] = 4 Next occurrence of t[2] = 50 PSC 2006

  43. Algorithm: Stage 1 1 2 3 4 5 6 7 8 9 50 50 100 50 50 50 50 100 50 next[1] = 2 next[2] = 4 next[3] = 8 Next occurrence of t[3] = 100 And so on… PSC 2006

  44. Algorithm: Stage 2 1 2 3 4 5 6 7 8 9 10 11 … 100 50 25 25 100 50 15 30 5 30 50 … S Consider an S at t[5]. PSC 2006

  45. Algorithm: Stage 2 1 2 3 4 5 6 7 8 9 10 11 … 100 50 25 25 100 50 15 30 5 30 50 … S Q PSC 2006

  46. Algorithm: Stage 2 1 2 3 4 5 6 7 8 9 10 11 … 100 50 25 25 100 50 15 30 5 30 50 … S Q Q PSC 2006

  47. Algorithm: Stage 2 1 2 3 4 5 6 7 8 9 10 11 … 100 50 25 25 100 50 15 30 5 30 50 … S 30+50 > 50 Q Q So Stop! PSC 2006

  48. Algorithm: Stage 2 1 2 3 4 5 6 7 8 9 10 11 … 100 50 25 25 100 50 15 30 5 30 50 … S Q Q Q PSC 2006

  49. Algorithm: Stage 2 1 2 3 4 5 6 7 8 9 10 11 … 100 50 25 25 100 50 15 30 5 30 50 … Q S Q Q Q So we get a new sequence t consisting of Q and S PSC 2006

  50. Algorithm: Stage 3 • Here we get t consisting of Q and S. • We first want to find the matches of r in t. • Define St  S in t, Qt  Q in t • Define Sr  S in r, Qr  Q in r PSC 2006

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